Problem 29 - Calculus with Concepts in Calculus

Question

Find the intervals on which the given Ir function is increasing and those on which it is decreasing. $$ f(x)=x^{2}+x+1 $$

Step-by-Step Solution

Answer

\( f(x) \) is decreasing on \((-\infty, -\frac{1}{2})\) and increasing on \((-\frac{1}{2}, \infty)\).

Step 1: Understand the Function

We have the function \( f(x) = x^2 + x + 1 \), which is a quadratic function. Quadratics are generally parabolas that open upwards when the coefficient of \( x^2 \) is positive.

Step 2: Find the Derivative

To determine where the function is increasing or decreasing, we need to find the derivative \( f'(x) \). The derivative \( f'(x) = \frac{d}{dx}(x^2 + x + 1) = 2x + 1 \).

Step 3: Determine Critical Points

Set the derivative equal to zero to find the critical points: \( 2x + 1 = 0 \). Solving this, we find \( x = -\frac{1}{2} \). This is the point where the function changes from increasing to decreasing or vice versa.

Step 4: Test Intervals Around Critical Point

Using the critical point, test intervals to see if \( f'(x) > 0 \) or \( f'(x) < 0 \).- For \( x < -\frac{1}{2} \), pick \( x = -1 \): \( f'(-1) = 2(-1) + 1 = -1 \) (negative, so \( f(x) \) is decreasing).- For \( x > -\frac{1}{2} \), pick \( x = 0 \): \( f'(0) = 2(0) + 1 = 1 \) (positive, so \( f(x) \) is increasing).

Step 5: State the Intervals

From the previous tests, we conclude:- The function \( f(x) = x^2 + x + 1 \) is **decreasing** on the interval \( (-\infty, -\frac{1}{2}) \).- The function is **increasing** on the interval \( (-\frac{1}{2}, \infty) \).

Key Concepts

Understanding Quadratic Functions

A quadratic function is one of the simplest types of polynomial functions that takes the form \( ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants and \( a eq 0 \). Quadratic functions produce a distinctive U-shaped graph known as a parabola.

Here are some key characteristics of quadratic functions that can help you understand them better:

• The term squared \( (x^2) \) makes the function nonlinear, setting it apart from linear functions.
• For positive \( a \), the parabola opens upwards. When \( a \) is negative, it opens downwards.
• The vertex of the parabola is the highest or lowest point, depending on the orientation.
• The line of symmetry goes through the vertex, dividing the parabola into two identical halves.

In the case of \( f(x) = x^2 + x + 1 \), the parabola opens upwards because the coefficient of \( x^2 \) is positive, and it doesn't intersect with the x-axis.

This means it has no real roots, indicating that the quadratic function never reaches zero.

Decoding the Derivative

The derivative of a function measures how the function's value changes as the input changes. It gives us the slope of the function at any given point.

When finding the intervals where a function is increasing or decreasing, we use its derivative:

• If the derivative is positive \( (f'(x) > 0) \), the function is increasing on that interval.
• If the derivative is negative \( (f'(x) < 0) \), the function is decreasing on that interval.

For the quadratic function \( f(x) = x^2 + x + 1 \), the derivative is calculated as:
\[ f'(x) = \frac{d}{dx}(x^2 + x + 1) = 2x + 1 \] This tells us that the slope of the function at any point \( x \) is \( 2x + 1 \). Testing around intervals of critical points can reveal whether the function is inclined upwards or downwards.

Identifying Critical Points

Critical points are where the derivative of a function equals zero or is undefined. These are the points at which the function may change its increasing or decreasing behavior.

To find critical points, we set the derivative equal to zero and solve for \( x \):

\[ 2x + 1 = 0 \]
\[ x = -\frac{1}{2} \]

This critical point \( x = -\frac{1}{2} \) is crucial as it helps in determining the behavior of the function on either side:

• If the function changes from increasing to decreasing, the point is a local maximum.
• If it changes from decreasing to increasing, the point is a local minimum.

For \( f(x) = x^2 + x + 1 \), the function decreases on \((-\infty, -\frac{1}{2})\) and increases on \((-\frac{1}{2}, \infty)\). This indicates that our critical point is a turning point where the slope changes from negative to positive, marking a transition from decreasing to increasing behavior.